limits - Evaluate $lim_{n to infty }frac{(n!)^{1/n}}{n}$.

Evaluate
$$\lim_{n \to \infty }\frac{(n!)^{1/n}}{n}.$$

Can anyone help me with this? I have no idea how to start with. Thank you.

Answer

Let's work it out elementarily by wisely applying Cauchy-d'Alembert criterion:

$$\lim_{n\to\infty} \frac{n!^{\frac{1}{n}}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{(n+1)}}\cdot \frac{n^{n}}{n!} = \lim_{n\to\infty} \frac{n^{n}}{(n+1)^{n}} =\lim_{n\to\infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}}=\frac{1}{e}. $$

Also notice that by applying Stolz–Cesàro theorem you get the celebre limit:

$$\lim_{n\to\infty} (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}} = \frac{1}{e}.$$

The sequence $L_{n} = (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}$ is called Lalescu sequence, after the name of a great Romanian mathematician, Traian Lalescu.

Q.E.D.